1. ## going in circles

Hi folks,

I am trying to solve the following question:

Find the equations of the two circles which pass through the point (2,0) and have both the y-axis and the line y - 1 = 0 as tangents.

This seems like an easy question because you can do it by scale drawing. The first (small) circle can't really go anywhere else - see attachment! But assuming you had to do it by calculation.

To find the equation of a circle we need 3 points on the circumference. We have one supplied directly (2,0) then we have the general equation:

$
x^2 + y^2 + 2fx + 2gy + c = 0 (1)
$

plus the two tangents:

x = 0

y = 1

substituting (2,0) gives:

4 + 4f + c = 0

and substituting x = 0 into the general equation gives:

$y^2 + 2gy -4(f + 1) = 0$

a quadtratic in y. Since the tangent touches the circle we expect a repeated root i.e. $b^2 - 4ac = 0$

$4g^2 + 4.1.4(f + 1) = 0$

$g^2 + 4f + 4 = 0 (4)$

then substitute y = 1 into the general equation:

$x^2 + 1 + 2fx + 2g -4(f + 1) = 0$

$x^2 + 2fx + 2g + 1 - 4(f + 1) = 0$

again setting $b^2 - 4ac = 0$ gives:

$
4f^2 + 16f - 8g + 12 = 0 (5)
$

Equations (4) and (5) are correct. What I mean is that when you substitue the correct answers in, they give consistent results but how do you solve them? It seems clear that I am on the wrong track. I then realised that the equation y = -x + 1 must be a diameter to both circles so I tried substituting this into the general equation (1) to solve for the intersection points. It yielded a result for x with many terms in f and g that I couldn't reduce to anything like the correct answer.

Any help here would be appreciated!
best regards

2. Hi

AS you said the line y=-x+1 is a diameter of the circles
The centers are on this line
Let C (xC,yC) one center of the possible circles
yC = -xC + 1

The distance between C and the point M(2,0) is such that CM² = (xC-2)²+yC² = 2xC² - 6 xC + 5
This distance is the same as the one between C and the y-axis, which is |xC|
Therefore : 2xC² - 6 xC + 5 = xC²
xC² - 6 xC + 5 = 0
(xC - 1)(xC - 5) = 0
xC = 1 or 5

3. Originally Posted by s_ingram
Hi folks,

I am trying to solve the following question:

Find the equations of the two circles which pass through the point 2,0) and have both the y-axis and the line y-1=0 as tangents.

...
1. I used this equation of a circle:

$(x-a)^2+(y-b)^2=r^2$ with the center at C(a, b) and the radius r.

2. You already know 3 points of the circle: (2, 0), (0, 1-r), (a, 1)

3. Substitute the coordinates of these points into the equation of the circle. You'll get a system of equations. Solve this system for a, b and r.

$\left|\begin{array}{rcl}(2-a)^2+(0-b)^2&=&r^2 \\ (0-a)^2+(a+1-b)^2&=&r^2 \\ (a-a)^2 + (1-b)^2&=&r^2 \end{array}\right.$

4. After some transformations and calculations you'll get:

Circle #1: $(1,0); r=1$
Circle #2: $(5,-4); r=5$

4. Hello, s_ingram!

Find the equations of the two circles which pass through the point (2,0)
and have both the $y$-axis and the line $y\,=\, 1$ as tangents.
Code:
        |
|         R
- + - P - * o * - - - y=1
----+---o-----:-----*----------
| * 2     :       *
|*        :        *
|         :
*         :         *
Q o - - - - o C       *
*       (x,y)       *
|
|*                 *
| *               *
|   *           *
|       * * *
|

The center of the circle is $C(x,y)$

It passes through the point $P(2,0)$
It is tangent to the $y$-axis at $Q(0,y)$
It is tangent to $y = 1$ at $R(x,1)$

The radius is: . $CP \:=\: CQ \:=\: CR \:=\: r$

We have: . $x \:=\:r$

Since $CR \:=\:1-y \:=\:r\:=\:x$, then: . $y \:=\:1-x$

Since $CP \,=\,R$, we have: . $\sqrt{(x-2)^2 + (y-0)^2} \:=\:r$

. . which becomes: . $\sqrt{(x-2)^2 + (1-x)^2} \:=\:x$

Then: . $(x-2)^2 + (1-x)^2 \:=\:x^2$

. . which simplifies to: . $x^2 - 6x + 5 \:=\:0$

. . which factors: . $(x-1)(x-5) \:=\:0$

. . and has roots: . $x \:=\:1,5$

. . . . . . . . Then: . $y \:=\:0,\text{-}4$

One circle has center $(1,0)$ and $r = 1\!:\quad \boxed{(x-1)^2 + y^2 \:=\:1}$

The other has center $(5,\text{-}4)$ and $r = 5\!:\quad \boxed{(x-5)^2 + (y+4)^2 \:=\:25}$

5. Excellent! Thanks a lot guys.